POJ1985 Cow Marathon树的直径

最新推荐文章于 2019-03-14 09:58:25 发布

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Farmer John希望为他的奶牛们规划一条马拉松路线，让他们得到更多的锻炼。本篇博客介绍了一个算法，用于找出农场地图上两个最远的点，以此作为马拉松的起点和终点，确保奶牛们能够获得最大程度的运动。

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http://poj.org/problem?id=1985

Description

After hearing about the epidemic of obesity in the USA, Farmer John wants his cows to get more exercise, so he has committed to create a bovine marathon for his cows to run. The marathon route will include a pair of farms and a path comprised of a sequence of roads between them. Since FJ wants the cows to get as much exercise as possible he wants to find the two farms on his map that are the farthest apart from each other (distance being measured in terms of total length of road on the path between the two farms). Help him determine the distances between this farthest pair of farms.

Input

* Lines 1.....: Same input format as "Navigation Nightmare".

Output

* Line 1: An integer giving the distance between the farthest pair of farms.

Sample Input

Sample Output

Hint

The longest marathon runs from farm 2 via roads 4, 1, 6 and 3 to farm 5 and is of length 20+3+13+9+7=52.

#include<iostream>
#define MAXN 1010000
#include<vector>
#include<cstdio>
#include<string.h>
using namespace std;
struct Edge
{
    int v,w;
    Edge(int vv,int ww):v(vv),w(ww) {}
};
int n,m;
int dist[MAXN],max_len,End;
vector<vector<Edge> >G;
void dfs(int u,int father,int len)
{
    //起点，父亲节点，两者之间长度
    if(len>max_len)max_len=len,End=u;//更新终点;
    for(int i=0; i<G[u].size(); i++)
    {
        //对于每一个节点，进行深搜;
        int v=G[u][i].v,w=G[u][i].w;//与之相连的点;
        if(v==father)continue;
        dist[v]=max(dist[v],len+w);//更新到子节点之间的最大距离;
        dfs(v,u,len+w);//继续深搜;
    }
}
int main()
{
    int u,v,w;
    char ch;
    cin>>n>>m;
    G.clear();//清空
    G.resize(n+2);//申请空间
    for(int i=1; i<=m; i++)
    {
        scanf("%d%d%d %c",&u,&v,&w,&ch);
        G[u].push_back(Edge(v,w));
        G[v].push_back(Edge(u,w));
        //无向图;
    }
    memset(dist,0,sizeof(dist));
    max_len=0;
    dfs(1,-1,0);//第一次深搜找到距离某点最远的点op
    dfs(End,-1,0);//求距离op最远的点，两次深搜求出直径;
    int ans=0;
    cout<<dist[End]<<endl;
    return 0;
}

#include<algorithm>
#include<iostream>
#include<cstdlib>
#include<cstring>
#include<cstdio>
#include<cmath>
#define LL long long
#define MOD 100000000
#define inf 2147483640
#define Pi acos(-1.0)
#define free(a) freopen(a".in","r",stdin),freopen(a".out","w",stdout);
using namespace std;
 
const int maxn=40010;
struct edge {int to,next,w;}e[maxn<<1];
int n,m,rt,ans,cnt,head[maxn];
 
void link(int u,int v,int w) {
    e[++cnt].to=v;e[cnt].next=head[u];head[u]=cnt;e[cnt].w=w;
    e[++cnt].to=u;e[cnt].next=head[v];head[v]=cnt;e[cnt].w=w;
}
void dfs(int x,int fa,int d) {
    if (ans<d) ans=d,rt=x;
    for (int i=head[x];i;i=e[i].next)
        if (e[i].to!=fa) dfs(e[i].to,x,d+e[i].w);
}
int main() {
    scanf("%d%d",&n,&m);
    char ch;
    for (int u,v,w,i=1;i<=m;i++) {
        scanf("%d%d%d %c",&u,&v,&w,&ch);
        link(u,v,w);
    }
    rt=ans=0;dfs(1,0,0);
    dfs(rt,0,0);
    printf("%d",ans);
    return 0;
}